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noma
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Boxes are left unchanged... but is what's inside them left unchanged?
Boxes are left unchanged regardless of what happens - all they are doing is looking in the box, they can't reorder them, boxes can't be removed, etc
The prisoners agree on what is the first, second, third, box etc. The first prisoner goes in. He opens boxes 1-50. (You didn't say he had to stop when he found his name: only that if he finds his name he gets taken to Room C.)
They're allowed to repeat this for 50 boxes - if they don't find their name, everyone dies. If they do, they're taken off to room C and have no contact with anyone else.
As he opens each box, he removes the paper inside. At the end he places all the papers in box 50. 50% success rate at this point.

2nd prisoner comes in (assuming prisoner 1 succeeded and everyone wasn't killed, of course.) He starts at box 50 -- as previously discussed by the group -- and opens boxes 50-99. If he finds his name, he leaves the papers in box 99, now holding 99 pieces of paper. Approx. 98% chance of finding his name.

If the 1st and 2nd prisoners succeeded, then the 3rd prisoner and all the rest now have a 100% success rate of finding their names, as prisoner number 3 will only need to open two boxes. He can then place all the papers in box 100 (although this is not necessary) and the rest of the prisoners only need to open one box and are also guaranteed to find their names.

Overall success percentage: 49% (if I've done my math right)

[Last edited by noma at 03-19-2008 02:12 AM]
03-19-2008 at 02:08 AM
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Rabscuttle
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That thought occurd to me too but that contradicts "all they are doing is looking in the boxes"

Maybe they should all just lie - 1dt guy "finds" his name in first box, 2nd guy "finds" his name in 2nd box etc :)
03-19-2008 at 02:49 AM
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coppro
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If the prisoners cannot communicate via any means once they enter Room B, the only way to ensure maximum success is to have each box opened the same number of times. So the first half of the prisoners should open the first half of the boxes, and the second half of the prisoners should open the second half of the boxes. There's no way to increase that if the prisoners can't move names (or to put odds of survival above 50% in any case).
03-19-2008 at 02:55 AM
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Rabscuttle
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Just gave up and looked up the answer. It's very neat.
03-19-2008 at 03:32 AM
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TripleM
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Theres no tricks at all (removing things from boxes, lying, etc). Theres a much better strategy than that Coppro, illogical as it seems.

Mathematical knowledge is required to come up with the rough probability of success, but even to non-mathematicians once told the strategy it shouldn't be too hard to why out why it works so well.
[Last edited by TripleM at 03-19-2008 04:26 AM]
03-19-2008 at 04:25 AM
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noma
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TripleM wrote:Mathematical knowledge is required to come up with the rough probability of success, but even to non-mathematicians once told the strategy it shouldn't be too hard to why out why it works so well.
Is mathematical knowledge required to work out the solution? If so, are we talking Newton or Grade 12 math?
[Last edited by noma at 03-19-2008 05:01 AM]
03-19-2008 at 04:57 AM
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TripleM
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Tough to say.. describing the strategy requires no mathematical knowledge at all. Realising why it's reasonable does somewhat.
03-19-2008 at 05:01 AM
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TripleM
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Time to begin with the subtle hints..

Step 1:
Click here to view the secret text
03-20-2008 at 12:10 AM
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Monkey
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This is a silly idea that could work if the first 2 prisoners have 50 sticky notes, the third has 1 sticky note, and all 3 have a pencil.

1st prisoner looks in the first 50 boxes and writes down on a sticky note what the name in the box is for each one. 50% getting his name.

2nd prisoner looks at the sticky notes, and if his name is on one of them, he takes it, and looks at another 49. Otherwise, he looks at the other 50. Either way, he does what prisoner 1 did if they don't have a sticky note. 100% getting his name.

3rd prisoner looks in the lone box, write down the name in it on sticky note, and scans through the sticky notes for his name. When he finds it, he takes it.

All the others have to do is find their name.

:D

note: not serious

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[Last edited by Monkey at 03-20-2008 12:37 AM]
03-20-2008 at 12:36 AM
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golfrman
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I think i got this, the prisoners line up in a certain order and give themselves numbers 1-100. They then give each of the boxes a number 1-100. number 1 goes in and picks the box with his number, # 1. If it has his number, he's fine. If it has another, he goes to that box. So if it's prisoner #5, he goes to box number 5. Then same thing until he's picked 50 boxes. Each of the prisoners then follow doing the same thing.

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03-20-2008 at 01:42 AM
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TripleM
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Correct. Anyone care to hazard a guess at why this gives a survival chance of over 30%?
[Last edited by TripleM at 03-20-2008 03:56 AM]
03-20-2008 at 03:56 AM
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bachus
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Ok. Here are my thoughts:

It can be observed that after the first prisoner enters and finds his name, the second prisoner finds his name if, starting from his box, he arrives at his name or at the first prisoner's name (that cycle is verified by the first prisoner, so it's sure it's safe).

The third prisoner has at each choice as valid boxes (that guarantee him succes) the boxes with the name of the first, the second and his own box (so 3 boxes take him for sure to his name)

..and so on

Now some maths:

Click here to view the secret text


[Last edited by bachus at 03-20-2008 01:49 PM : inserted image]
03-20-2008 at 01:48 PM
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noma
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Well, it's non-random and I suppose any non-random picking raises the odds. Although pretty good at basic math, I can't go further than that.

Of course, even if the prisoners succeed, they will still all be executed. It's only logical.

Proof follows:
Click here to view the secret text

QED ;)
[Last edited by noma at 03-20-2008 01:58 PM]
03-20-2008 at 01:51 PM
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krammer
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bachus wrote:
It can be observed that after the first prisoner enters and finds his name, the second prisoner finds his name if, starting from his box, he arrives at his name or at the first prisoner's name (that cycle is verified by the first prisoner, so it's sure it's safe).
Huh? How does it help at all if the second prisoner finds the first's name?

Also, the strategy doesn't explain how to deal with loops. What does a prisoner do if redirected to a box he has already seen? Or do we assume they just go for the next box in sequence?

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03-20-2008 at 02:58 PM
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bachus
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There can be no loops for a prisoners, if he starts from his box it is sure that after a number of boxes he will find his name. The question is if he finds it in 50 tries.

A loop would mean that he can arrive at a box from 2 different boxes, which is false, as the names in the boxes are unique.

Take the following configuration:

boxes:
1 2 3 4 5 6
names
3 1 2 4 6 5

you can see that the first prisoner goes 1 - 3 - 2 and finds his name

the second prisoner goes 2 (sees 1, which means he will find his name) - 1 - 3 (where indeed he finds it)

The explanation is that there are closed groups, above {1, 2, 3}, {4} and {5, 6}, and in a group you are guaranteed to go sequentially to all the names. That's why, if 2 sees 1's name , he knows they are in the same group and therefore he will find his own name.
03-20-2008 at 03:11 PM
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krammer
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Ugh. I really should have seen that. Sorry, I wasn't thinking clearly.

But still, the 2nd prisoner finding the 1st name doesn't guarantee finding his in enough boxes - as you say, it only guarantees he will eventually get there.

Mathematically, then:
So, writing the members of the symmetry group S100 as cycles, we need that for all n, a)number n is found in the cycle also containing position n and b)they are less than 50 apart (in the correct direction).

If a is true, then b) will always be true if the cycle is less than 50 long, and if it is longer then there is a 50/L chance it is, where L is cycle length. So b) can at worst halve the probability of a).

Calculating a) on the other hand, is I think beyond me. I'm pretty sure your formula is wrong though. Particularly when you say that if the first 3 find their numbers, the rest are guaranteed to. Suppose 1,2 and 3 are in their own box, but then 4 contains 5, 5 contains 6, etc, up to 100 containing 4. Then the first 3 find their name but nobody else does...

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03-20-2008 at 04:10 PM
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stigant
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I imagine he's waiting to be rewarded for getting the right answer.
Well, I certainly appreciate the mod points, but I was actually waiting for confirmation, inspiration, and then finally access to post a puzzle (I've been on vacation this week, and so not on the boards). As I have only 2 of the three and we seem to have moved on...

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03-21-2008 at 02:41 AM
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Rabscuttle
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krammer wrote:
Ugh. I really should have seen that. Sorry, I wasn't thinking clearly.

But still, the 2nd prisoner finding the 1st name doesn't guarantee finding his in enough boxes - as you say, it only guarantees he will eventually get there.

He will find it in exactly the same number of boxes as the first. Because they start at their own box, and the number they are looking for points back to the box they start with, they always take the length of the loop they are on to find their own name.

eg 5 -> 3 -> 2 -> 10 -> 1
10 -> 1 -> 5 -> 3 -> 2
[Last edited by Rabscuttle at 03-21-2008 02:53 AM]
03-21-2008 at 02:52 AM
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zex20913
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For krammer's part a), we can say that the paper in the box says "go to box #____". The cycle must end with "go to box # (Prisoner's number)."

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03-21-2008 at 06:35 AM
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bachus
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Yes, the prisoner will make the cycle always in the [cyclelength], which the previous prisoner has verified to be valid.

But your example, krammer, is very good and it proves also that my calculations are wrong. If before a prisoner i there was one cycle of length 1, then that name is not a valid option anymore (so it cannot have the probability 1/n of appearing, it is 0). I stand by my formula (just the input is wrong), the right way to do it would be to say that the valid boxes at each extraction for prisoner i are the names of the prisoners in the same cycle with him. But I don't have any clue of how to express this mathematically...
03-21-2008 at 08:13 AM
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golfrman
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So, the next puzzle.
You are in a 100 story building, and you have two of an object, and you want to see at what floor this object will first break. What is the minimal amount of drops you would have to maketo see when the object would break?

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03-24-2008 at 01:23 AM
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TripleM
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Just to give a full (mathematical) explanation of the solution to the last one, for those who are mathematically inclined:

Click here to view the secret text


Its been proven apparently the prisoners can't do any better than this, but I wouldn't have a clue how to go about that :P

[Last edited by TripleM at 03-24-2008 02:18 AM]
03-24-2008 at 02:17 AM
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Dolan42
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golfrman wrote:
So, the next puzzle.
You are in a 100 story building, and you have two of an object, and you want to see at what floor this object will first break. What is the minimal amount of drops you would have to maketo see when the object would break?
14 drops.
Start at floor 14 and drop it, if it breaks drop the other from 1-13 for 13+1 drops max.
If it didn't break, move up 13 floors, if it breaks drop from 15-26 for 12+2 drops max.
If it didn't break there move up 11 floors, then 10 then 9 etc.

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03-24-2008 at 04:43 AM
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RoboBob3000
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Zero drops. It will always break at the ground floor.

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03-24-2008 at 06:12 AM
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TripleM
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RoboBob3000 wrote:
Zero drops. It will always break at the ground floor.

Not if you drop it 0 times :P
03-24-2008 at 07:10 AM
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bachus
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I guess this one with the eggs has been here already :) ... but... perhaps one should not read the entire topic searching for duplicates and simply let the people who don't know the answer, answer.

Anyway, I think dolan got it, so he should post another.
03-24-2008 at 12:05 PM
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golfrman
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Yeah, he got it, maybe i should've looked to see if that was there before posting it.

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03-25-2008 at 11:10 PM
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Dolan42
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So, Beethro is out exploring a small tropical island with a small village. After a few days searching for a dungeon and not finding one he decides to leave. As the locals gather to say good bye he makes a small speech, during the speech he comments "I have enjoyed meeting all of you. I feel that we have much in common and some differences, for example I have brown eyes and all of you have either blue or green eyes, but..." at that point the entire village got up and got very angry and Beethro had to run for his life to his boat to escape. What he didn't realize is that the villagers had prefect logic and an odd custom. If any villager found out the color of their eyes they would have to commit suicide the following morning. Obviously mirrors were very rare and telling someone their eye color to kill them usually ended up with them telling you yours. 3 weeks later Beethro returned in the afternoon and found 73 fresh corpses and several new graves. There was nobody left alive. How many villagers were there originally and why?



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[Last edited by Dolan42 at 03-27-2008 06:27 AM : typo]
03-27-2008 at 06:14 AM
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TripleM
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Click here to view the secret text

[Last edited by TripleM at 03-27-2008 06:50 AM]
03-27-2008 at 06:49 AM
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Dolan42
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Yep, your go.

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03-27-2008 at 02:08 PM
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